Evaluate $~~\int^e_1\dfrac{-2\ln x}{x^2}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $ -2$ (Choice B) B $ \dfrac2e-2$ (Choice C) C $ \dfrac4e-2$ (Choice D) D $ \dfrac4e+2$
We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv= \dfrac1{x^2}\,dx\,$. Then $~du = \dfrac1x\,dx~$ and $~v = -\dfrac1x\,$. Integration by parts gives $ -2\int^e_1 \dfrac{\ln x}{x^2}\,dx =-2\Big(-\dfrac{\ln x}x\Bigg]_1^e+\int^e_1\dfrac1{x^2}dx\Big)$ $ \,=\dfrac{2\ln x}x+\dfrac2x\Bigg]^e_1 $ $ \,=\Big(\dfrac2e+\dfrac2e\Big)-\Big(0+2\Big) = \dfrac4e-2$